Derivative of exponential function: (a^x) lim(h->0)(a^h-1)/h

[serenaleesl]

The proof of the derivative of a^x using limit approach: -

(a^x)*Lim (h->0) (a^h - 1 )/ h
= (a^x)*ln a.

how does the limit approach gives rise to ln a?

 

[mmm4444bot]

Re: Derivative of exponential function

Hello Serena:

That's an excellent question; I have no idea what the answer is.

Of course, you left out several steps leading up to what you posted, but I'll assume that's because you understand the limit definition of the derivative of f(x) = a[sup:2kraljb7]x[/sup:2kraljb7] up to that point.

Hopefully (so that I don't have to type it all out :wink: ), you also understand that the limit you've posted is equal to f'(0).

So, we can see that f'(x) = f'(0) * a^x.

At this point, we clearly need to know the derivative in order to determine the derivative! In other words, we're stuck.

I believe that introductory calculus courses would, at this point, simply define f'(0) to be ln(a), and you're off to the races.

If you were familiar with the chain rule (I'm assuming not, since these limit definitions generally come well before teaching the chain rule), then I would derive the ln(a) result for you another way, and that would provide an indirect definition for the limit that you posted. But, it would still be indirect; it would not show how to evaluate this limit.

It's a wild guess, but my gut feeling is that much higher levels of analysis (perhaps something to do with Taylor Polynomials) is required to evaluate this limit directly. (I'm confident that somebody else will chime in regarding this point, and, in the event that my entire post is off-base, perhaps an evaluation of the limit will be posted.)

I realize that my response to your question is less than satisfactory, but that's the way introductory mathematics goes. Some axioms* must simply be provided until later skills and knowledge are available.

If you were working the same issue with f(x) = e[sup:2kraljb7]x[/sup:2kraljb7], then your limit would be exactly the same, except that a[sup:2kraljb7]h[/sup:2kraljb7] would become e[sup:2kraljb7]h[/sup:2kraljb7]. And, again, your course would most likely define this limit to be equal to 1 without showing how.

Cheers,

~ Mark

*A proposition that is assumed or provided without proof for the sake of studying the consequences that follow from it.

 

[PAULK]

Re: Derivative of exponential function

The proof of the derivative of a^x using limit approach: -
(a^x)*Lim (h->0) (a^h - 1 )/ h
= (a^x)*ln a.

how does the limit approach gives rise to ln a?

For this, you need this more basic limit:

e^(x + h) - e^x
lim ---------------- = e^x
h->0 h

I.e. prove that the derivative of e^x is e^x.

{Sorry, I don't have a nice proof of that for you. I think I saw one about 50 years ago in my elementary calculus book but...}

Now you want to compute:

a^(x + h) - a^x
lim ----------------
h->0 h

Remembering that a^x can be written e^(x ln a):

(e^ln a)^(x + h) - (e^ln a)^x
lim ------------------------------
h->0 h

e^((x + h)ln a) - e^(x ln a)
lim ---------------------------------
h->0 h

e^(x ln a + h ln a) - e^(x ln a)
lim ---------------------------------
h->0 h

Substitute: k = h ln a, and y = x ln a

e^(y + k) - e^y
lim --------------------
h->0 h/ln a

e^(y + k) - e^y
ln a lim ----------------------
k->0 k

ln a e^y =

ln a e^(x ln a) =

ln a a^x

 

[Deleted member 4993]

Re: Derivative of exponential function

I.e. prove that the derivative of e^x is e^x.

{Sorry, I don't have a nice proof of that for you. I think I saw one about 50 years ago in my elementary calculus book but...}

I learned that statement as the definition of e^x (that is a function whose derivative is itself).

 

[mmm4444bot]

Re: Derivative of exponential function

For this, you need this more basic limit ...

Hi Paulk:

I may have misinterpreted Serena's question, but I read it as asking how to evaluate the posted limit versus how to show that d/dx a^x is a^x * ln(a).

In other words, how would one evaulate limits of the following type as strictly an exercise in limit evaluation?

\(\displaystyle \lim_{h \to 0} \frac{a^h - 1}{h}\)

\(\displaystyle \lim_{h \to 0} \frac{e^h - 1}{h}\)

I.e. prove that the derivative of e^x is e^x.

{Sorry, I don't have a nice proof of that for you ...

I've seen this one:

\(\displaystyle y \;=\; e^x\)

\(\displaystyle ln(y) \;=\; x\)

\(\displaystyle \frac{1}{y} \cdot \frac{dy}{dx} \;=\; 1\)

\(\displaystyle \frac{dy}{dx} \;=\; y\)

\(\displaystyle \frac{dy}{dx} \;=\; e^x\)

Cheers,

~ Mark

 

[galactus]

Re: Derivative of exponential function

Here is an approach I use:

\(\displaystyle \lim_{h\to 0}\frac{a^{x+h}-a^{x}}{h}\)

\(\displaystyle \lim_{h\to 0}\frac{a^{x}a^{h}-a^{x}}{h}\)

\(\displaystyle \lim_{h\to 0}\frac{a^{x}(a^{h}-1)}{h}\)

Since a^x does not have h, we can factor it out:

\(\displaystyle f'(x)=a^{x}\lim_{h\to 0}\frac{a^{h}-1}{h}\)

Now, this is a famous limit we know equals ln(a) and we have \(\displaystyle a^{x}ln(a)\)

As for evaluating \(\displaystyle \lim_{h\to 0}\frac{e^{h}-1}{h}=1\)

We can do this by expressing the derivative of e^x at x=0 as a limit:

\(\displaystyle \frac{d}{dx}[e^{x}]\) at x=0.

\(\displaystyle \lim_{h\to 0}\frac{e^{0+h}-e^{0}}{h}=\lim_{h\to 0}\frac{e^{h}-1}{h}=1\)

 

[galactus]

Re: Derivative of exponential function

Here is another way to evaluate this limit I just figured out.

\(\displaystyle \lim_{h\to 0}\frac{e^{h}-1}{h}\)

Let \(\displaystyle 1+\frac{1}{x}=e^{h}\)

\(\displaystyle h=ln(1+\frac{1}{x})\)

Make the subs:

\(\displaystyle \lim_{x\to {\infty}}\frac{1}{xln(1+\frac{1}{x})}\)

\(\displaystyle \lim_{x\to {\infty}}\frac{1}{ln(1+\frac{1}{x})^{x}}\)

Notice what is in the ln?. That is the famous e limit.

So we have \(\displaystyle \frac{1}{ln(e)}=1\)

 

[mmm4444bot]

Re: Derivative of exponential function

...

\(\displaystyle f'(x)=a^{x}\lim_{h\to 0}\frac{a^{h}-1}{h}\)

Now, this is a famous limit we know equals ln(a)

Hi Galactus!

The question in my mind remains ... how do we know that the expression goes to ln(a) as h goes to zero?

Cheers,

~ Mark

 

[mmm4444bot]

Re: Derivative of exponential function

... Make the subs:

\(\displaystyle \lim_{x\to {\infty}}\frac{1}{xln(1+\frac{1}{x})}\)

I understand the substitutions, except for how h->0 changes to x->infinity.

What is the justification for this switch?

I spent about 45 minutes with the Great God Google, but I could not locate an answer to my question; the evaluations of this type of limit that I found justify the value of the limit by introducing something extra (eg: famous limits).

My gut feeling remains that, strictly as an exercise in evaluating limits, there's no simple analysis -- at least nothing involving straightforward algebra.

Cheers,

~ Mark

 

[galactus]

Re: Derivative of exponential function

The justification for the limit change to infinity is simply as \(\displaystyle h\rightarrow 0^{+}, \;\ then \;\ x\rightarrow {\infty}^{+}\)

and \(\displaystyle h\rightarrow 0^{-}, \;\ then \;\ x\rightarrow{\infty}^{-}\)

For the limit \(\displaystyle \lim_{h\to 0}\frac{a^{h}-1}{h}=ln(a)\), we can cheat and use L'Hopital. But that wouldn't be the way to go.

Let's use the series for \(\displaystyle \frac{a^{h}-1}{h}\)

It is:

\(\displaystyle \lim_{h\to 0}\sum_{k=0}^{\infty}\frac{h^{k}(ln(a))^{k+1}}{(k+1)!}=\lim_{h\to 0}\left[ln(a)+\frac{h(ln(a))^{2}}{2}+\frac{h^{2}(ln(a))^{3}}{3!}+\frac{h^{3}(ln(a))^{4}}{4!}+..............\right]\)



Now, see what we have left because all the h's cancel out. Just ln(a)

Yes, I am sure there is a slick algebraic way. there always is if we can find it. Let me look at yet another way and get back.

 

[galactus]

Re: Derivative of exponential function

Here we go. Another way using something similar to what I done before.

\(\displaystyle \lim_{h\to 0}\frac{a^{h}-1}{h}\)

Let \(\displaystyle a^{h}=1+\frac{1}{x}\)

\(\displaystyle ln(1+\frac{1}{x})=hln(a)\)

\(\displaystyle h=\frac{ln(1+\frac{1}{x})}{ln(a)}\)

Make the subs and we get:

\(\displaystyle \lim_{x\to {\infty}}\frac{1}{\frac{xln(1+\frac{1}{x})}{ln(a)}}\)

\(\displaystyle \lim_{x\to {\infty}}\frac{ln(a)}{ln(\underbrace{1+\frac{1}{x})^{x}}_{\text{this is e}}}\)

The denominator is ln(e)=1 due to the famous e limit once again.

We have \(\displaystyle \boxed{\frac{ln(a)}{1}=ln(a)}\).........TA...DAA

 

[mmm4444bot]

Re: Derivative of exponential function

The justification for the limit change to infinity is simply as \(\displaystyle h\rightarrow 0^{+}, \;\ then \;\ x\rightarrow {\infty}^{+}\)

and \(\displaystyle h\rightarrow 0^{-}, \;\ then \;\ x\rightarrow{\infty}^{-}\)

Hi Galactus:

Oh, of course! (I was looking at e^h, DOH.)

I like your latest post, but, again, it relies on infamy. :wink:

The series you posted is also interesting, but I've forgotten enough that I'll need to save it for some future date when I'm caught up. (My eight-year math sabbatical left me quite rusty in some areas.)

~ Mark

 

[galactus]

Re: Derivative of exponential function

Oh well, if you're a purist and require a proof of \(\displaystyle \lim_{x\to {\infty}}(1+\frac{1}{x})^{x}=e\) that can be arranged. But I am not that much of a purist.

That would be like proving \(\displaystyle \lim_{x\to 0}\frac{sin(x)}{x}=1\) everytime we use it. :wink:

Assuming that is what you meant by infamy.

 

[mmm4444bot]

Re: Derivative of exponential function

... That would be like proving \(\displaystyle \lim_{x\to 0}\frac{sin(x)}{x}=1\) everytime we use it. :wink:

Point taken!

Really, I am pleased with your last example. It's accessible at an introductory-calculus level. I added it to my notebook!

Thank you,

~ Mark

 

[PAULK]

Re: Derivative of exponential function

Here is another way to evaluate this limit I just figured out.

\(\displaystyle \lim_{h\to 0}\frac{e^{h}-1}{h}\)

Let \(\displaystyle 1+\frac{1}{x}=e^{h}\)

\(\displaystyle h=ln(1+\frac{1}{x})\)

Make the subs:

\(\displaystyle \lim_{x\to {\infty}}\frac{1}{xln(1+\frac{1}{x})}\)

\(\displaystyle \lim_{x\to {\infty}}\frac{1}{ln(1+\frac{1}{x})^{x}}\)

Notice what is in the ln?. That is the famous e limit.

So we have \(\displaystyle \frac{1}{ln(e)}=1\)

Very nice! That's it!
I am not sure I saw this 50 years ago, but it will do.

Of course, all of this is unnecessary if we just say e^x is the inverse of ln(x), and we don't have to mess with limits. That is an easy way out, and I was hoping to avoid that.

Now the logic goes:

lim (1 + 1/x)^x exists and is some number, WHICH WE SHALL CALL e.
x->inf

Then the above shows that e^x has itself as the derivative.

Good work!

 

[galactus]

Re: Derivative of exponential function

Thanks for the kudos . I derived that myself . I did not get it from a book. If it is similar, just a coincidence.

I like your observation about the derivative.